An algebra problem by ابراهيم فقرا

Since we are given values of $f(x)$ at consecuive points, we can use the Method of Differences here.

$\begin{array}{cccccc} x & f(x) & D_1 & D_2 & D_3 & D_4 & D_5 \\ { \begin{array}{c} 1 \\ 2 \\ 3 \\ 4 \\ 5 \\ 6 \end{array} } & { \begin{array}{c} 2 \\ 2 \\ 3 \\ 4 \\ 6 \\ 9 \end{array} } & { \begin{array}{c} 0 \\ 1 \\ 1 \\ 2 \\ 3 \end{array} } & { \begin{array}{c} 1 \\ 0 \\ 1 \\ 1 \end{array} } & { \begin{array}{c} -1 \\ 1 \\ 0 \end{array} } & { \begin{array}{c} 2 \\ -1 \end{array} } & { \begin{array}{c} -3 \end{array} } \end{array}$

$f(x)$ is a quintic polynomomial, so ( D_5 ) must be constant. Then, we can continue the table and calculate all differences backwards to get $f(7)$ .

$\begin{array}{cccccc} x & f(x) & D_1 & D_2 & D_3 & D_4 & D_5 \\ { \begin{array}{c} 1 \\ 2 \\ 3 \\ 4 \\ 5 \\ 6 \\ 7 \end{array} } & { \begin{array}{c} 2 \\ 2 \\ 3 \\ 4 \\ 6 \\ 9 \\ {\color{#D61F06}\boxed{9}} \end{array} } & { \begin{array}{c} 0 \\ 1 \\ 1 \\ 2 \\ 3 \\ {\color{#D61F06}0} \end{array} } & { \begin{array}{c} 1 \\ 0 \\ 1 \\ 1 \\ {\color{#D61F06}-3} \end{array} } & { \begin{array}{c} -1 \\ 1 \\ 0 \\ {\color{#D61F06}-4} \end{array} } & { \begin{array}{c} 2 \\ -1 \\ {\color{#D61F06}-4} \end{array} } & { \begin{array}{c} -3 \\ {\color{#D61F06}-3} \end{array} } \end{array}$

Let $f(x)=ax^5+bx^4+cx^3+dx^2+ex+f$

So $f(1)=a+b+c+d+e+f=2$

$f(2)=32a+16b+8c+4d+2e+f=2$

$f(3)=243a+81b+27c+9d+3e+f=3$

$f(4)=1024a+256b+64c+16d+4e+f=4$

$f(5)=3125a+625b+125c+25d+5e+f=6$

$f(6)=7776a+1296b+216c+36d+6e+f=9$

If you solve this system of equations above, you will get that

$f(x)=\frac{-3*x^5+55*x^4-375*x^3+1205*x^2-1722*x+1080}{120}$

$f(7)=9$